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(D) The length of the wire is $l = \pi r$,where $r$ is the radius of the semicircle.Therefore,$r = \frac{l}{\pi}$.Since the wire is a thin semicircle,

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$\frac{2 m l^2}{\pi^2}$

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(D) The length of the wire is $l = \pi r$,where $r$ is the radius of the semicircle.Therefore,$r = \frac{l}{\pi}$.Since the wire is a thin semicircle,every point on the wire is at a distance $r$ from the center of the semicircle.The moment of inertia of the entire wire about an axis passing through the center of the semicircle and perpendicular to its plane is $I_{center} = \int r^2 dm = r^2 \int dm = m r^2$.However,the question asks for the moment of inertia about an axis passing through the end of the wire and perpendicular to its plane.Since all mass elements $dm$ are at a distance $r$ from the center,but the axis is shifted to the end,we must consider the distance of each point from the end.For a semicircle,the moment of inertia about an axis passing through one end and perpendicular to the plane is $I = \int r^2 dm$. Since every point on the wire is at a distance $r$ from the center,but the axis is at the end,we use the parallel axis theorem or direct integration. For a thin wire bent into a semicircle,the moment of inertia about an axis perpendicular to the plane passing through one end is $I = 2mr^2$.Substituting $r = \frac{l}{\pi}$,we get $I = 2m(\frac{l}{\pi})^2 = \frac{2ml^2}{\pi^2}$.

$A$ thin wire of length $l$ and mass $m$ is bent in the form of a semicircle. The moment of inertia about an axis perpendicular to its plane and passing through the end of the wire is .............

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